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I've been trying to optimize the ridge regression loss function by gradient descent, but I've been banging my head against it for a while. The loss function can be written as

$$L(w,b)=\sum_{i=1}^n(y^{(i)}-(w\cdot x^{(i)}+b))^2+\lambda\|w\|^2_2$$

with $w$ being the vector of parameters, and $b$ is the intercept. Naively I would simply differentiate wrt. $w$ and try to minimize that function;

$$\frac{dL}{dw}=-2\sum_{i=1}^n(y^{(i)}-(w\cdot x^{(i)}+b))x^{(i)}+2\lambda\|w\|$$

Now, my initial idea would be to assimilate the intercept into the parameter vector in the loss function, and differentiate wrt. my new $\tilde{w}=(b,w)$. But if doing so, wouldn't this mean also changing the regularization term, and thus add a penalty on the intercept? It's my understanding that this should be avoided. How do I assimilate the intercept without this resulting in a penalty also on the intercept value in the regularization term?

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I'd suggest assimilating via $v = (b, w)$ and doing $$ L(v) = \|y - Xv\|^2 + \lambda v^T \Omega v $$ where $$ \Omega = \text{diag}\left(0, 1, \dots, 1\right) $$ so $$ v^T\Omega v = \sum_{j \geq 2} v_j^2 = w^T w. $$

Now we have $$ L(v) = y^Ty - 2v^TX^Ty + v^T(X^TX + \lambda \Omega)v $$ so $$ \nabla L = -2X^Ty + 2 (X^TX + \lambda \Omega)v $$ and then $$ \nabla L \stackrel{\text{set}}= 0 \implies (X^TX + \lambda \Omega)v = X^Ty. $$


Regarding invertibility of $X^TX + \lambda \Omega$, let $z \in \mathbb R^p \backslash \{0\}$ and consider $$ z^T (X^TX + \lambda \Omega)z. $$ Let's try to make this $0$ to see if $X^TX + \lambda \Omega$ is PSD (positive semidefinite) or PD (positive definite). $z^T \Omega z = \sum_{j\geq 2}z_j^2$ so the only way to get $\lambda z^T \Omega z = 0$ is to have $z \propto e_1$, the first standard basis vector. So WLOG take $z = e_1$. Then $$ e_1^TX^TXe_1 = \|Xe_1\|^2 = \|X_1\|^2 $$ where $X_1$ is the first column of $X$. So this means that $X^TX + \lambda \Omega$ will be properly PD and not just PSD exactly when the first column of $X$, which we've made our unpenalized column by our choice of $\Omega$, is "full rank", which in this case just means non-zero. That's easy to check (just look at your data) so we don't have to worry about non-invertibility even though we're not doing a full ridge regression.


One slightly unsatisfactory part of this penalty is that it makes the Bayesian interpretation less nice because to recover this as a MAP estimate we'd need $$ v\sim \mathcal N\left(0, \frac{\sigma^2}\lambda \Omega^{-1}\right) $$ except $\Omega$ is singular, so we don't actually have a pdf for this (so e.g. if you wanted to choose $\lambda$ by maximizing the marginal likelihood as with a Gaussian process, now you can't, or at least you won't get a Gaussian likelihood since it's not defined). One possible way to get this back is to instead view this as a mixed model $$ y = b\mathbf 1 + Zw + \varepsilon $$ where $w\sim \mathcal N(0, \frac{\sigma^2}{\lambda} I)$ now plays the role of a random effect, and $Z$ is $X$ without the first intercept-giving column. The actual estimation process may change here depending on how much you want this to actually be a mixed model but I think it's an interesting alternative. In particular, we can now integrate $w$ out and still end up with a valid Gaussian likelihood, and in the spirit of REML we could integrate $b$ out too if we put an improper uniform prior on it.

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  • $\begingroup$ +1 on making the point of concatenating $b$ and $w$, and define the $\Omega$ in the way that do not regularize the intercept. $\endgroup$
    – Haitao Du
    May 15 '18 at 18:44
  • $\begingroup$ Very nice way of not penalizing the intercept! I've tried to run through the math to follow what's going on, and unless I'm mistaken, in the expression for $\nabla L$, shouldn't the second term be positive? As in $\nabla L = -2X^Ty+2(X^TX+\lambda\Omega)v$ $\endgroup$
    – AstroAT
    May 15 '18 at 19:59
  • $\begingroup$ @AndersT Thanks, and yeah you're right, that was a sign error. Fixed now $\endgroup$
    – jld
    May 15 '18 at 20:02

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